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Greeley [361]
3 years ago
7

At standard temperature and pressure, carbon dioxide has a density of 1.98 kg/m3. What volume does 1.70 kg of carbon dioxide occ

upy at standard temperature and pressure?
Physics
1 answer:
nikitadnepr [17]3 years ago
8 0

Answer:

<h2>volume= 0.85m^3</h2>

Explanation:

<em>The density of a substance is defined as the mass per unit volume of the substance, the unit is in kg/m^3 and it is represented by the greek letter rho</em>

Step one:

given data

we are told that the density  of Co2=  1.98 kg/m3

and the mass of Co2 is= 1.70 kg

we know the relation between mass, volume and density is

density=mass/volume

make volume subject of formula we have

volume=mass/density

substitute we have

volume=1.7/1.98\\\\volume= 0.85m^3

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3 years ago
A Ferris wheel has radius 5.0 m and makes one revolution every 8.0 s with uniform rotation. A person who normally weighs 670 N i
DaniilM [7]

Answer:

The apparent weight of the person as she pass the highest point is  N  =  458.8 \ N

Explanation:

From the question we are told that

   The radius of the Ferris wheel is r = 5.0 \ m

    The period of revolution is T = 8.0 \ s

     The weight of the person is  W  =  670 \ N

   

Generally the speed of the wheel is mathematically represented as

      v =  \frac{2 \pi r}{T }

substituting values

      v =  \frac{2 * 3.142 *  5}{8 }

       v =  3.9 3 \ m/s

The apparent weight (the normal force exerted on her by the bench) at the highest point is mathematically evaluated as

          N  =  mg  - \frac{mv^2}{r}

Where m is the mass of the person which is mathematically evaluated as

     m =  \frac{W}{g}

substituting values

    m =  \frac{670}{9.8}

    m =  68.37 \ kg

So

    N  =  68.37 * 9.8   - \frac{68.37 * {3.93}^2}{5}

    N  =  458.8 \ N

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3 years ago
Cassy shoots a large marble (Marble A, mass: 0.06kg) at a smaller marble (Marble B, mass: 0.03kg) that is sitting still. Marble
4vir4ik [10]

The question can be solved using conservation of linear momentum.

M_{a} = 0.06kg and M_{b} = 0.03kg

Let the initial velocity of Marble A be , V_{a1} = 0.7m/s

Let the initial velocity of Marble B be, V_{b1} = 0m/s

Let the velocity of Marble A after collisiong , V_{a2}= -0,02m/s

Let the velocity of Marble B after collision be V_{b2}

From the conservation of linear momentum equation. We get,

M_{a}V_{a1}+M_{b}V_{b1}=M_{a}V_{a2}+M_{b}V_{b2}

Substituting the values we get,

(0.06)(0.7) + 0 = (0.06)(-0.02) + (0.03)V_{b2}

we get, V_{b2} = 1.44m/s

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3 years ago
Why does a tail appear when a comet nears the sun?
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